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CSCE 3110 Data Structures & Algorithm Analysis. Stacks and Queues Reading: Chap.3 Weiss. Stacks. Stack: what is it? ADT Applications Implementation(s). What is a stack?. Stores a set of elements in a particular order Stack principle: LAST IN FIRST OUT = LIFO
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CSCE 3110Data Structures & Algorithm Analysis Stacks and Queues Reading: Chap.3 Weiss
Stacks • Stack: what is it? • ADT • Applications • Implementation(s)
What is a stack? • Stores a set of elements in a particular order • Stack principle: LAST IN FIRST OUT • = LIFO • It means: the last element inserted is the first one to be removed • Example • Which is the first element to pick up?
Last In First Out top B A C B A D C B A E D C B A D C B A top top top top A
Stack Applications • Real life • Pile of books • Plate trays • More applications related to computer science • Program execution stack (read more from your text) • Evaluating expressions
objects: a finite ordered list with zero or more elements.methods: for all stack Stack, item element, max_stack_size positive integerStack createS(max_stack_size) ::= create an empty stack whose maximum size is max_stack_size Boolean isFull(stack, max_stack_size) ::= if (number of elements in stack == max_stack_size) return TRUEelse return FALSEStack push(stack, item) ::=if (IsFull(stack)) stack_fullelse insert item into top of stack and return Stack ADT
Boolean isEmpty(stack) ::= if(stack == CreateS(max_stack_size)) return TRUEelse return FALSEElement pop(stack) ::= if(IsEmpty(stack)) returnelse remove and return the item on the top of the stack. Stack ADT (cont’d)
Array-based Stack Implementation • Allocate an array of some size (pre-defined) • Maximum N elements in stack • Bottom stack element stored at element 0 • last index in the array is the top • Increment top when one element is pushed, decrement after pop
Stack createS(max_stack_size) ::= #define MAX_STACK_SIZE 100 /* maximum stack size */ typedef struct { int key; /* other fields */ } element; element stack[MAX_STACK_SIZE]; int top = -1;BooleanisEmpty(Stack) ::= top< 0;BooleanisFull(Stack) ::= top >= MAX_STACK_SIZE-1; Stack Implementation: CreateS, isEmpty, isFull
void push(int *top, element item){ /* add an item to the global stack */ if (*top >= MAX_STACK_SIZE-1) { stack_full( ); return; } stack[++*top] = item;} Push
element pop(int *top){ /* return the top element from the stack */ if (*top == -1) return stack_empty( ); /* returns and error key */ return stack[(*top)--]; } Pop
void push(pnode top, element item){ /* add an element to the top of the stack */ pnode temp = (pnode) malloc (sizeof (node)); if (IS_FULL(temp)) { fprintf(stderr, “ The memory is full\n”); exit(1); } temp->item = item; temp->next= top; top= temp; } List-based Stack Implementation: Push
element pop(pnode top) {/* delete an element from the stack */ pnode temp = top; element item; if (IS_EMPTY(temp)) { fprintf(stderr, “The stack is empty\n”); exit(1); } item = temp->item; top = temp->next; free(temp); return item;} Pop
Algorithm Analysis • push O(?) • pop O(?) • isEmpty O(?) • isFull O(?) • What if top is stored at the beginning of the array?
A LegendThe Towers of Hanoi • In the great temple of Brahma in Benares, on a brass plate under the dome that marks the center of the world, there are 64 disks of pure gold that the priests carry one at a time between these diamond needles according to Brahma's immutable law: No disk may be placed on a smaller disk. In the begging of the world all 64 disks formed the Tower of Brahma on one needle. Now, however, the process of transfer of the tower from one needle to another is in mid course. When the last disk is finally in place, once again forming the Tower of Brahma but on a different needle, then will come the end of the world and all will turn to dust.
The Towers of HanoiA Stack-based Application • GIVEN: three poles • a set of discs on the first pole, discs of different sizes, the smallest discs at the top • GOAL: move all the discs from the left pole to the right one. • CONDITIONS: only one disc may be moved at a time. • A disc can be placed either on an empty pole or on top of a larger disc.
Towers of Hanoi – Recursive Solution void hanoi (int discs, Stack fromPole, Stack toPole, Stack aux) { Disc d; if( discs >= 1) { hanoi(discs-1, fromPole, aux, toPole); d = fromPole.pop(); toPole.push(d); hanoi(discs-1,aux, toPole, fromPole); }
Applications • Infix to Postfix conversion [Evaluation of Expressions]
Is the End of the World Approaching? • Problem complexity 2n • 64 gold discs • Given 1 move a second
X = a / b - c + d * e - a * ca = 4, b = c = 2, d = e = 3Interpretation 1: ((4/2)-2)+(3*3)-(4*2)=0 + 8+9=1Interpretation 2:(4/(2-2+3))*(3-4)*2=(4/3)*(-1)*2=-2.66666…How to generate the machine instructions corresponding to a given expression? precedence rule + associative rule Evaluation of Expressions
user compiler Postfix: no parentheses, no precedence
#define MAX_STACK_SIZE 100 /* maximum stack size */#define MAX_EXPR_SIZE 100 /* max size of expression */typedef enum{1paran, rparen, plus, minus, times, divide, mod, eos, operand} precedence;int stack[MAX_STACK_SIZE]; /* global stack */char expr[MAX_EXPR_SIZE]; /* input string */ Infix to Postfix Assumptions: operators: +, -, *, /, % operands: single digit integer
Evaluation of Postfix Expressions int eval(void){/* evaluate a postfix expression, expr, maintained as a global variable, ‘\0’ is the the end of the expression. The stack and top of the stack are global variables. get_token is used to return the token type and the character symbol. Operands are assumed to be single character digits */ precedence token; char symbol; int op1, op2; int n = 0; /* counter for the expression string */ int top = -1; token = get_token(&symbol, &n); while (token != eos) { if (token == operand) push(&top, symbol-’0’); /* stack insert */
else { /* remove two operands, perform operation, and return result to the stack */ op2 = pop(&top); /* stack delete */ op1 = pop(&top); switch(token) { case plus: push(&top, op1+op2); break; case minus: push(&top, op1-op2); break; case times: push(&top, op1*op2); break; case divide: push(&top, op1/op2); break; case mod: push(&top, op1%op2); } } token = get_token (&symbol, &n); } return pop(&top); /* return result */}
precedence get_token(char *symbol, int *n){/* get the next token, symbol is the character representation, which is returned, the token is represented by its enumerated value, which is returned in the function name */ *symbol =expr[(*n)++]; switch (*symbol) { case ‘(‘ : return lparen; case ’)’ : return rparen; case ‘+’: return plus; case ‘-’ : return minus;
case ‘/’ : return divide; case ‘*’ : return times; case ‘%’ : return mod; case ‘\0‘ : return eos; default : return operand; /* no error checking, default is operand */ }}
Infix to Postfix Conversion(Intuitive Algorithm) (1) Fully parenthesized expression a / b - c + d * e - a * c --> ((((a / b) - c) + (d * e)) – (a * c)) (2) All operators replace their corresponding right parentheses. ((((a / b) - c) + (d * e)) – (a * c)) (3) Delete all parentheses. ab/c-de*+ac*- two passes - * * / + -
The orders of operands in infix and postfix are the same. a + b * c, * > +
a *1 (b +c) *2 d match ) *1 = *2
Rules (1) Operators are taken out of the stack as long as their in-stack precedence is higher than or equal to the incoming precedence of the new operator. (2) ( has low in-stack precedence, and high incoming precedence. ( ) + - * / % eos isp 0 19 12 12 13 13 13 0 icp 20 19 12 12 13 13 13 0
precedence stack[MAX_STACK_SIZE];/* isp and icp arrays -- index is value of precedencelparen, rparen, plus, minus, times, divide, mod, eos */static int isp [ ] = {0, 19, 12, 12, 13, 13, 13, 0};static int icp [ ] = {20, 19, 12, 12, 13, 13, 13, 0}; isp: in-stack precedence icp: incoming precedence
Infix to Postfix void postfix(void){/* output the postfix of the expression. The expression string, the stack, and top are global */ char symbol; precedence token; int n = 0; int top = 0; /* place eos on stack */ stack[0] = eos; for (token = get _token(&symbol, &n); token != eos; token = get_token(&symbol, &n)) { if (token == operand) printf (“%c”, symbol); else if (token == rparen ){
Infix to Postfix (cont’d) /*unstack tokens until left parenthesis */ while (stack[top] != lparen) print_token(delete(&top)); pop(&top); /*discard the left parenthesis */ } else{ /* remove and print symbols whose isp is greater than or equal to the current token’s icp */ while(isp[stack[top]] >= icp[token] ) print_token(delete(&top)); push(&top, token); } } while ((token = pop(&top)) != eos) print_token(token); print(“\n”);}
Queue • Stores a set of elements in a particular order • Stack principle: FIRST IN FIRST OUT • = FIFO • It means: the first element inserted is the first one to be removed • Example • The first one in line is the first one to be served cinemark
Queue Applications • Real life examples • Waiting in line • Waiting on hold for tech support • Applications related to Computer Science • Threads • Job scheduling (e.g. Round-Robin algorithm for CPU allocation)
First In First Out D C B rear front A B A C B A D C B A rear front rear front rear front rear front
objects: a finite ordered list with zero or more elements.methods: for all queue Queue, item element, max_ queue_ size positive integerQueue createQ(max_queue_size) ::= create an empty queue whose maximum size ismax_queue_sizeBoolean isFullQ(queue, max_queue_size) ::= if(number of elements in queue == max_queue_size) returnTRUEelse returnFALSEQueue Enqueue(queue, item) ::= if (IsFullQ(queue)) queue_full else insert item at rear of queue and return queue Queue ADT